Power-Sum

Digital root of $2^{1}.=2$

Digital root of $2^{2}.=4$

Digital root of $2^{3}.=8$

Digital root of $2^{4}.=1+6=7$

Digital root of $2^{5}.=3+2=5$

Digital root of $2^{6}.=6+4=10=1+0=1$

Digital root of $2^{7}.=2$

Digital root of $2^{8}.=4$

Digital root of $2^{9}.=8$

Digital root of $2^{10}.=1+0+2+4=7$ And so on.........

So we observe that the cycle repeats after every 7 powers of 2.

Dividing 100 by 7,we observe that the remainder is 4

So the digital root of Digital root of $2^{100}$ .=7

It is a well known fact that $dr(ab) \equiv dr(a) \times dr(b) \pmod 9$

It follows that $dr(a^b) \equiv dr(a)^b \pmod 9$

So, $dr(2^{100}) \equiv dr(1024)^{10} \equiv 7^{10} \equiv 7^{10 \pmod {\phi (9)}} \equiv 7^4 \equiv 7 \pmod 9$

Thus the answer is $\boxed{7}$

Who said Agnishom cannot do math without computers?